A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge at point P, w

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A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge at point P, which is on the x-axis, 9.83 mm from the origin

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Thu Thảo 3 years 2021-08-18T21:03:57+00:00 1 Answers 142 views 0

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  1. thuthuy
    0
    2021-08-18T21:05:12+00:00
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    Answer:

    E = (-3.61^i+1.02^j) N/C

    magnitude E = 3.75N/C

    Explanation:

    In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

    \vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

    Where the minus sign means that the electric field point to the charge.

    k: Coulomb’s constant = 8.98*10^9Nm^2/C^2

    q = -4.28 pC = -4.28*10^-12C

    r: distance to the charge from the point P

    The point P is at the point (0,9.83mm)

    θ: angle between the electric field vector and the x-axis

    The angle is calculated as follow:

    \theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

    The distance r is:

    r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

    You replace the values of all parameters in the equation (1):

    \vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

    The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

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